Question

Find the points on the curve $y = \sqrt {x - 3} $ where the tangent is perpendicular to the line $6x + 3y - 5 = 0$.

Answer 1

Hint: Firstly, find the slope of the line. Thereafter, differentiate the curve with respect to \[n\].
The slope of line is given by, $y = mx + c$.


Complete step by step solution:
Given curve, $y = \sqrt {x - 3} $
And given line $ = 6x + 3y - 5 = 0$
$6x + 3y - 5 = 0$
$3y = 5 - 6x$
$y = \dfrac{{5 - 6x}}{3}$
$y = \dfrac{5}{3} - \dfrac{6}{3}x$
$y = \dfrac{5}{3} - 2x$
$y = - 2x + \dfrac{5}{3}$
We compare this line by $y = mx + c$then,
Slope \[\left( {{m_1}} \right)\]$ = 2$
Now, $y = \sqrt {x - 3} $
$y = {(x - 3)^{\dfrac{1}{2}}}$
Differentiate with respect to$x$, we will get
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}{(x - 3)^{\dfrac{1}{2} - 1}}$ .
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}{(x = 3)^{\dfrac{1}{2} - 1}}$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}{(x - 3)^{\dfrac{{1 - 2}}{2}}}$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}{(x - 3)^{ - \dfrac{1}{2}}}$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}{\left( {\sqrt {x - 3} } \right)^{ - 1}}$
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {x - 3} }}$ $....(1)$
We know that, when two lines are perpendicular, then the slopes are negative reciprocal.
Then \[{m_1} \times {m_2} = - 1\]
If ${m_1} = - 2$then
$( - 2) \times {m_2} = - 1$
$({m_2}) = \dfrac{{ + 1}}{{ + 2}}$
$({m_2}) = \dfrac{1}{2}$ $....(2)$
We will substitute equation (2) in equation (1), we have
$\dfrac{1}{2} = \dfrac{1}{{2\sqrt {x - 3} }}$
Now, cross multiplying the values, we will get
$2\sqrt {x - 3} = 2$
$\sqrt {x - 3} = 1$
Squaring both sides, we get
${\sqrt {\left( {x - 3} \right)} ^2} = {(1)^2}$
$x - 3 = 1$
$x = 4$
Now, $y = \sqrt {x - 3} $ $....(3)$
Put the value of $x = 4$in equation (3), we have
$y = \sqrt {4 - 3} $
$y = \sqrt 1 $
$y = 1$
Therefore, the points on the curve is $(4,1)$


Note: When solving these types of problems,remember the result which says two slopes are perpendicular to each other then their product is $ - 1$.